Orbital tetrahedral bond

The concepts of directed valence and orbital hybridization were developed by Linus Pauling soon after the description of the hydrogen molecule by the valence bond theory. These concepts were applied to an issue of specific concern to organic chemistry, the tetrahedral orientation of the bonds to tetracoordinate carbon. Pauling reasoned that because covalent bonds require mutual overlap of orbitals, stronger bonds would result from better overlap. Orbitals that possess directional properties, such as p orbitals, should therefore be more effective than spherically symmetric 5 orbitals. 

The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer. When an 5 orbital hybridizes rvith three p orbitals, the resultant sp3 hybrid orbitals are unsyimmetrical about the nucleus. One of the two... 

Pig. 1. Polar graph showing the dependence on and q> of a tetrahedral bond orbital. The value of y> in cross-section is shown the function is cylindrically symmetrical about the midline of this cross-section. 

In this discussion of the transition elements we have considered only the orbitals (n— )d ns np. It seems probable that in some metals use is made also of the nd orbitals in bond formation. In gray tin, with the diamond structure, the four orbitals 5s5p3 are used with four outer electrons in the formation of tetrahedral bonds, the 4d shell being filled with ten electrons. The structure of white tin, in which each atom has six nearest neighbors (four at 3.016A and two at 3.17.5A), becomes reasonable if it is assumed that one of the 4d electrons is promoted to the 5d shell, and that six bonds are formed with use of the orbitals 4dSs5p35d. 

After rising at copper and zinc, the curve of metallic radii approaches those of the normal covalent radii and tetrahedral covalent radii (which themselves differ for arsenic, selenium, and bromine because of the difference in character of the bond orbitals, which approximate p orbitals for normal covalent bonds and sp3 orbitals for tetrahedral bonds). The bond orbitals for gallium are expected to be composed of 0.22 d orbital, one s orbital, and 2.22 p orbitals, and hence to be only slightly stronger than tetrahedral bonds, as is indicated by the fact that R(l) is smaller than the tetrahedral radius. 

Although we have described the structures of several molecules in terms of hybrid orbitals and VSEPR, not all structures are this simple. The structures of H20 (bond angle 104.4°) and NH3 (bond angles 107.1°) were described in terms of sp3 hybridization of orbitals on the central atom and comparatively small deviations from the ideal bond angle of 109° 28 caused by the effects of unshared pairs of electrons. If we consider the structures of H2S and PH3 in those terms, we have a problem. The reason is that the bond angle for H2S is 92.3°, and the bond angles in PH3 are 93.7°. Clearly, there is more than a minor deviation from the expected tetrahedral bond angle of 109° 28 caused by the effect of unshared pairs of electrons ... 

Although Chapter 25 does not address directly why some compounds with coordination 4 are tetrahedral and some are square planar, it is possible to surmise that the answer lies with (1) Crystal Field Theory and the energies of the d orbitals involved bonding and (2) how many unpaired electrons the metal complex has. 

If nitrogen uses only its p orbitals in bond formation, the angle between N-H bonds would be 90°. However, compounds prefer formations in which electrons are as far apart as possible. For ammonia this is made possible by forming a tetrahedral structure in which the angle between the bonds (M-H) is 107°. This is only possible by undergoing sp3 hybridization. 

The formation of a Si crystal is shown in Fig. 1.10. Aside from the core, each Si atom has four valence electrons two 3s electrons and two 3p electrons. To form a Si crystal, one of the 3s electrons is excited to the 3p orbital. The four valence electrons form four sp hybrid orbitals, each points to a vertex of a tetrahedron, as shown in Fig. 1.10. Thpse four sp orbitals are unpaired, that is, each orbital is occupied by one electron. Since the electron has spin, each orbital can be occupied by two electrons with opposite spins. To satisfy this, each of the directional sp orbitals is bonded with an sp orbital of a neighboring Si atom to form electron pairs, or a valence bond. Such a valence bonding of all Si atoms in a crystal form a structure shown in (b) of Fig. 1.10, the so-called diamond structure. As seen, it is a cubic crystal. Because all those tetrahedral orbitals are fully occupied, there is no free electron. Thus, similar to diamond, silicon is not a metal. 

Answer. Nitrogen undergoes sp hybridization, not sp, so it is tetrahedral. The additional sp orbital is occupied by a lone pair of electrons from the nitrogen. This lone pair results in electron-electron repulsion that causes the other sp orbitals bonded to fluorines to be closer together than the normal 109° tetrahedral bond angle, hence the 107.3 F-N-F bond angle. 

Fig. 4-2.—The angular dependence of a tetrahedral orbital with bond direction along the x axis.

Derivation of Results about Tetrahedral Orbitals.—The results about tetrahedral bond orbitals described. above are derived in the following way. We assume that the radial parts of the wave functions fa and faxf faM are so closely similar that their differences can be neglected. The angular parts are... 

An equivalent set of tetrahedral bond orbitals, differing from the e only in orientation, is... 

Since tetrahedral orbitals form stronger bonds than other orbitals it might be thought that hybridization to tetrahedral orbitals would always occur in bond formation. The tendency to use the best bond orbitals is, however, resisted in the case of atoms with an unshared pair (or more than one) by the tendency to keep the unshared pair in the 8 orbital, which is more stable than the p orbitals. In OFi, for example, the use of tetrahedral orbitals in bond formation would require that half of the 8 orbital (which is divided equally among the four tetrahedral orbitals) be used for shared pairs and only half for unshared pairs. Since a shared pair counts as only one electron for each atom, this would involve the loss of one-quarter of the extra stability due to a pair of s electrons, and the atom will strive to prevent this. On the... 

Fig. 4-6.—Tetrahedral bond orbital with 4 percent d character and 2 percent /.

With this structure the nickel atom lias achieved the krypton electron configuration its outer shell contains five unshared pairs (in the five M orbitals) and five shared pairs (occupying the 4s4p3 tetrahedral bond orbitals). The Ni—C bond length expected for this structure is about 2.16 A, as found by use of the tetrahedral radius 1.39 A obtained by extrapolation from the adjacent values in Table 7-13 (Cu, 1.35 A Zn, 1.31 A). 

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